Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
std::vector<long long> G[100001];
std::vector<long long> psn[100001];
std::vector<long long> arr[100001][20];
long long par[100001], lvl[100001];
long long P[100001][20];
void dfs_visit(long long s, long long p) {
  if (p != -1) {
    par[s] = p;
    lvl[s] = 1 + lvl[p];
  }
  for (long long i = 0, j = (long long)G[s].size(); i < j; i++) {
    if (G[s][i] != p) dfs_visit(G[s][i], s);
  }
  return;
}
std::vector<long long> merger(std::vector<long long> left,
                              std::vector<long long> right) {
  std::vector<long long> res;
  long long n1 = (long long)left.size(), n2 = (long long)right.size();
  long long i = 0, j = 0, k = 0;
  while (i != n1 && j != n2 && k != 10) {
    if (left[i] < right[j]) {
      res.push_back(left[i]);
      i++;
    } else if (left[i] > right[j]) {
      res.push_back(right[j]);
      j++;
    } else {
      res.push_back(left[i]);
      i++;
      j++;
    }
    k++;
  }
  while (i != n1 && k != 10) {
    res.push_back(left[i]);
    i++;
    k++;
  }
  while (j != n2 && k != 10) {
    res.push_back(right[j]);
    j++;
    k++;
  }
  return res;
}
void ST_build(long long n) {
  for (long long i = 1; i <= n; i++) {
    for (long long j = 0; (1 << j) <= n; j++) P[i][j] = -1;
  }
  for (long long i = 1; i <= n; i++) {
    P[i][0] = par[i];
    arr[i][0] = merger(psn[i], psn[par[i]]);
  }
  for (long long j = 1; (1 << j) <= n; j++) {
    for (long long i = 1; i <= n; i++) {
      if (P[i][j - 1] != -1) {
        P[i][j] = P[P[i][j - 1]][j - 1];
        arr[i][j] = merger(arr[i][j - 1], arr[P[i][j - 1]][j - 1]);
      }
    }
  }
  return;
}
std::vector<long long> LCA_find(long long p, long long q) {
  std::vector<long long> res = merger(psn[p], psn[q]);
  if (lvl[p] < lvl[q]) {
    long long tmp = p;
    p = q;
    q = tmp;
  }
  long long lx = 0;
  for (lx = 1; (1 << lx) <= lvl[p]; lx++)
    ;
  lx--;
  for (long long i = lx; i >= 0; i--) {
    if ((lvl[p] - (1 << i)) >= lvl[q]) {
      res = merger(res, arr[p][i]);
      p = P[p][i];
    }
  }
  if (p == q) return res;
  for (long long i = lx; i >= 0; i--) {
    if ((P[p][i] != -1) && (P[p][i] != P[q][i])) {
      res = merger(res, arr[p][i]);
      res = merger(res, arr[q][i]);
      p = P[p][i];
      q = P[q][i];
    }
  }
  res = merger(res, arr[p][0]);
  res = merger(res, arr[q][0]);
  return res;
}
int main(void) {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  std::cout.tie(NULL);
  long long n, m, q;
  std::cin >> n >> m >> q;
  for (long long i = 0; i < n - 1; i++) {
    long long s, d;
    std::cin >> s >> d;
    G[s].push_back(d);
    G[d].push_back(s);
  }
  for (long long i = 1; i <= m; i++) {
    long long t;
    std::cin >> t;
    if ((long long)psn[t].size() < 10) psn[t].push_back(i);
  }
  dfs_visit(1, -1);
  ST_build(n);
  while (q--) {
    long long v, u, a;
    std::cin >> v >> u >> a;
    std::vector<long long> ans = LCA_find(v, u);
    a = std::min(a, (long long)ans.size());
    std::cout << a << " ";
    for (long long i = 0; i < a; i++) std::cout << ans[i] << " ";
    std::cout << "\n";
  }
  return 0;
}